Question

(a) At the end of Example 25.4, it was stated that the critical angle for light going from diamond to air is 24.4º. Verify this.

(b) What is the critical angle for light going from zircon to air?

Short answer

1. As a result, as light travels from the diamond to the atmosphere, the critical angle is $24.4^{°}$.
2. As a result, the critical angle for zircon light to reach the atmosphere is$31.{33}^{°}$

Step-by-step solution

Definition of the critical angle

The critical angle is the angle of incidence that produces a 90-degree angle of refraction.

The critical angle for light going grom diamond to air is 24.4°.

a)

Consider the given information.

The expression to find the critical angle is,

${\theta }_{\circ }={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Where, n₁is the denser medium's index of refraction while the less dense medium's index of refraction is n₂.

For diamonds, the refractive index is 2.419. Then,

$$\begin{gathered} {\theta }_{\circ }={\sin }^{-1}\left(\frac{1}{2.419}\right) \\ =24.4^{°} \end{gathered}$$

As a result, the critical angle for light passing through a diamond to reach the atmosphere is $24.4^{°}$. It means for the angle of incidence greater than $24.4^{°}$In the diamond, the light is completely internally reflected. If the incidence angle is less than $24.4^{°}$, the light ray is refracted as it travels through the air.

Therefore, the critical angle for light traveling from the diamond to the atmosphere is $24.4^{°}$.

Calculating critical angle for light going from zircon to air

b)

Let us calculate the critical angle for light going from zircon to air.

The Refractive index of zircon is:$n_1=1.923$.

The expression to find the critical angle is,

${\theta }_{\circ }={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Substituting the given values, we get

$$\begin{gathered} {\theta }_{\circ }={\sin }^{-1}\left(\frac{1}{1.923}\right) \\ =31.{33}^{°} \end{gathered}$$

Therefore, the critical angle for light passing through zircon to reach the atmosphere is$31.{33}^{°}$.